Upper bound of the sum $\sum_{i=2}^{N}{\frac{1}{\log(i)}}$ One of the questions in Sierpinski's book on number theory lead to finding a tight upper bound for the following sum:
$$\sum_{i=2}^N {\frac{1}{\log(i)}}$$
The trivial upper bound like $\frac{N-1}{\log(2)}$ wouldn't work. Can someone please suggest a stronger bound?
 A: Given a decreasing positive function, in your case $f(x) = \frac{1}{\log x},$ it is standard to compare with the integral of the same thing. This is done by, in essence, starting at any integer $i$ and the point $(i,f(i))$ and drawing a horizontal segment of length exactly $1$ to the right. This segment is higher than the graph of the real-valued function. Eventually you get to the right endpoint, your $N,$ and that final segment extends over the graph out to $N+1.$ So
$$  \sum_{i=2}^N f(i) \; > \; \int_2^{N+1} \; f(x) dx.   $$
An antiderivative is $\mbox{li} \; x,$ see LOGARITHMIC INTEGRAL A pretty good lower bound is
$$   \mbox{li} \; (N+1) - \mbox{li} \; 2   $$ 
There is a similar process for an upper bound. Draw segments to the left. In case, as here, $f(1)$ is undefined, just start the process one later and keep the explicit $f(2)$ term. 
$$  \sum_{i=2}^N f(i) \; < \; f(2) +  \; \int_2^{N} \; f(x) dx.   $$
 A pretty good upper bound is
$$ \frac{1}{\log 2} +  \mbox{li} \; N - \mbox{li} \; 2  $$ 
There are any number of ways to discuss the size of $\mbox{li} \; x,$ see SUM. One exact sum I like is
$$  \mbox{li} \; x = \gamma + \log \log x + \sum_{n=1}^\infty \; \frac{(\log x)^n}{n \; n!}.   $$
